Question

Suppose (a) \(f\) is continuous for \(x \geq 0\), (b) \(f^{\prime}(x)\) exists for \(x>0\), (c) \(f(0)=0\), (d) \(f^{\prime}\) is monotonically increasing. Put $$ g(x)=\frac{f(x)}{x} \quad(x>0) $$ and prove that \(g\) is monotonically increasing.

Short answer

Question: Prove that the function \(g(x) = \frac{f(x)}{x}\) is monotonically increasing, given the conditions (a) \(f\) is continuous for \(x \geq 0\), (b) \(f^{\prime}(x)\) exists for \(x>0\), (c) \(f(0)=0\), and (d) \(f^{\prime}\) is monotonically increasing. Answer: We use the given conditions and the Mean Value Theorem to show that \(g^{\prime}(x) > 0\) for all \(x>0\). This proves that the function \(g(x)\) is monotonically increasing on \((0, +\infty)\).

Step-by-step solution

Find \(g^{\prime}(x)\).

To prove that the function \(g(x)\) is monotonically increasing, we need to find its derivative \(g^{\prime}(x)\) and show that it is positive for all \(x>0\). We will start by using the Quotient Rule for differentiation, where for a function \(g(x) = \frac{f(x)}{x}\), we have: $$ g^{\prime}(x) = \frac{f^{\prime}(x) \cdot x - f(x) \cdot 1}{x^2} $$

Analyze \(g^{\prime}(x)\) under given conditions.

The derivative of \(g(x)\), \(g^{\prime}(x)\), is: $$ g^{\prime}(x) = \frac{x f^{\prime}(x) - f(x)}{x^2} $$ Now we can analyze this expression for the given conditions. Since \(f^{\prime}\) is monotonically increasing, this means that for any pair of numbers \(x_1\) and \(x_2\) such that \(0 < x_1 < x_2\), we have: $$ f^{\prime}(x_1) < f^{\prime}(x_2) $$ We want to prove that there exists a similar inequality for \(g^{\prime}(x)\) that ensures that \(g^{\prime}(x) > 0\) for all \(x>0\).

Show that \(g^{\prime}(x)>0\) for all \(x>0\) using the Mean Value Theorem.

We will use the Mean Value Theorem (MVT) to demonstrate that \(g^{\prime}(x)>0\) for all \(x>0\). According to the MVT, if \(f(x)\) is continuous on a closed interval \([x_1, x_2]\) and differentiable on the open interval \((x_1, x_2)\), then there exists a value \(c\) in the interval \((x_1, x_2)\) such that: $$ f^{\prime}(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1} $$ Applying this to our function \(f\) and the conditions provided, we find that there exists a value \(c\) between \(x_1\) and \(x_2\), such that: $$ f^{\prime}(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1} $$ Now consider the inequality \(f^{\prime}(x_1) < f^{\prime}(x_2)\). Multiplying both sides by \(x_2 - x_1 > 0\), we obtain: $$ (x_2 - x_1)f^{\prime}(x_1) < (x_2 - x_1)f^{\prime}(x_2) $$ That is, $$ x_2f^{\prime}(x_1) - x_1f^{\prime}(x_1) < x_2f^{\prime}(x_2) - x_1f^{\prime}(x_2) $$ Taking the equation \(f^{\prime}(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\) from the MVT and substituting it into the inequality above, we get: $$ f(x_2) - f(x_1) < x_2f^{\prime}(x_2) - x_1f^{\prime}(x_2) $$ Dividing both sides by \(x_1x_2(x_2-x_1)\), we obtain: $$ \frac{f(x_2)-f(x_1)}{x_1x_2} < g^{\prime}(x_2) - g^{\prime}(x_1) $$ Since the left side of the inequality is positive, we can conclude that \(g^{\prime}(x_2) > g^{\prime}(x_1)\) for any \(x_1, x_2\) with \(0<x_1<x_2\). This implies that \(g(x)\) is monotonically increasing on \((0, +\infty)\).

Key concepts

Continuous Function

A continuous function is one that is smooth and unbroken—meaning the graph of the function can be drawn without lifting the pen from the paper. When we say a function like f(x) is continuous for x ≥ 0, we're describing a situation where there are no sudden jumps or gaps in the values of the function for any non-negative real numbers.

Measuring continuity at a point means checking three conditions: the function is defined at that point, the limit exists at that point, and the function's value equals that limit. The benefit of continuous functions is their predictability and the fact that we can apply calculus principles such as the Mean Value Theorem without worrying about irregularities.

To visualize, imagine driving down a smoothly paved road without any potholes or speed bumps—that's akin to moving along a continuous function.

Derivative Exists

When we say the derivative exists for a function, we mean at each point in a specified domain, there's a slope or a rate of change that we can calculate. This derivative tells us how fast or slow the function is changing at any given point.

For example, if we're talking about a car's speed as a function of time, the derivative would give us the exact acceleration of the car at every moment. When given that f'(x) exists for x > 0, it confirms that for all positive values of x, there's a meaningful slope. This consistency allows us to analyze how the function is increasing or decreasing, which is invaluable in proving that another function, like g(x), is monotonically increasing.

Mean Value Theorem

The Mean Value Theorem (MVT) is a cornerstone of calculus, providing a guarantee that, under the right conditions, a continuous and differentiable function will have at least one point where the instantaneous rate of change (the derivative) matches the average rate of change over an interval.

More formally, if a function f is continuous on [a, b] and differentiable on (a, b), then there's at least one point c in (a, b) where f'(c) equals the average slope between a and b (i.e., f'(c) = (f(b) - f(a)) / (b - a)).

It's like saying if you've traveled from point A to point B, there must have been at least one moment where you were traveling at the average speed of your entire trip. We use the MVT in various proofs and analyses to draw conclusions about function behavior.

Quotient Rule for Differentiation

When dealing with the division of two functions, the Quotient Rule for Differentiation comes into play. It provides a method for finding the derivative of a quotient of two differentiable functions.

The rule states that if you have a function g(x) = f(x) / h(x), then the derivative g'(x) can be found using the formula: g'(x) = (f'(x)h(x) - f(x)h'(x)) / [h(x)]^2.

Similar to how we deal with fractions in basic arithmetic, the numerator is differentiated, and the denominator's differentiation is subtracted, all over the square of the denominator. This rule is essential for finding the rate of change of ratios and comes into significant use for problems involving proportions like the one in our exercise.